Use information from Appendix \(D\) to calculate the pH of (a) a solution that is \(0.250 \mathrm{M}\) in sodium formate \((\mathrm{HCOONa})\) and \(0.100 M\) in formic acid \((\mathrm{HCOOH}) ;\) (b) a solution that is \(0.510 \mathrm{M}\) in pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) and \(0.450 \mathrm{M}\) in pyridinium chloride \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right) ;\) (c) a solution that is made by combining \(55 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) hydrofluoric acid with \(125 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium fluoride.

The concentrations of the acid (formic acid, \(\mathrm{HCOOH}\)) and its conjugate base (sodium formate, \(\mathrm{HCOONa}\)) are given as \(0.150\,\mathrm{M}\) and \(0.250\,\mathrm{M}\), respectively.

For a buffer solution, we can use the Henderson-Hasselbalch equation, which is given as follows: \(pH = pK_a + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) According to Appendix D, the \(pK_a\) value for formic acid is 3.75. Now we can substitute the values into the equation: \(pH = 3.75 + \log_{10} \frac{0.250}{0.150}\)

Now we can perform the calculation: \(pH = 3.75 + \log_{10} \frac{0.250}{0.150} \approx 4.24\) Thus, the pH of solution (a) is approximately 4.24. ###Solution (b)###

The concentrations of the base (pyridine, \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\)) and its conjugate acid (pyridinium chloride, \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NHCl}\)) are given as \(0.490\,\mathrm{M}\) and \(0.450\,\mathrm{M}\), respectively.

For a buffer solution, we can use the Henderson-Hasselbalch equation. Since this is a basic buffer solution, we need to use the \(pK_b\) value for pyridine. According to Appendix D, the \(pK_b\) value for pyridine is 8.75. First, we need to find the \(pK_a\) value using the relationship: \(pK_a + pK_b = 14\) \(pK_a = 14 - pK_b = 14 - 8.75 = 5.25\) Now we can substitute the values into the equation: \(pH = 5.25 + \log_{10} \frac{0.490}{0.450}\)

Now we can perform the calculation: \(pH = 5.25 + \log_{10} \frac{0.490}{0.450} \approx 5.40\) Thus, the pH of solution (b) is approximately 5.40. ###Solution (c)###

Since we are given the volumes and concentrations of the solutions, we can first calculate the moles of the acid (\(\mathrm{HF}\)) and its conjugate base (\(\mathrm{F}^{-}\)): moles of \(\mathrm{HF} = 0.050\,\mathrm{M} \times 0.080\,\mathrm{L} = 0.004\,\mathrm{mol}\) moles of \(\mathrm{F}^{-} = 0.10\,\mathrm{M} \times 0.100\,\mathrm{L} = 0.010\,\mathrm{mol}\) Now that we have the moles, we need to find the concentrations in the final mixed solution. To do this, we first calculate the total volume of the solution: \(V_{\text{total}} = 80\,\mathrm{mL} + 100\,\mathrm{mL} = 180\,\mathrm{mL} = 0.180\,\mathrm{L}\) Now we can calculate the concentrations: \([\mathrm{HF}] = \frac{0.004\,\mathrm{mol}}{0.180\,\mathrm{L}} = 0.0222\,\mathrm{M}\) \([\mathrm{F}^{-}] = \frac{0.010\,\mathrm{mol}}{0.180\,\mathrm{L}} = 0.0556\,\mathrm{M}\)

Now that we have the concentrations, we can use the Henderson-Hasselbalch equation. According to Appendix D, the \(pK_a\) value for hydrofluoric acid is 3.17: \(pH = 3.17 + \log_{10} \frac{0.0556}{0.0222}\)

Now we can perform the calculation: \(pH = 3.17 + \log_{10} \frac{0.0556}{0.0222} \approx 3.78\) Thus, the pH of solution (c) is approximately 3.78.